BZOJ1179或洛谷3672 [APIO2009]抢掠计划

BZOJ原题链接

洛谷原题链接

在一个强连通分量里的\(ATM\)机显然都可被抢，所以先用\(tarjan\)找强连通分量并缩点，在缩点的后的\(DAG\)上跑最长路，然后扫一遍酒吧记录答案即可。

#include<cstdio>
using namespace std;
const int N = 5e5 + 10;
struct eg{
	int x, y;
};
eg a[N];
int fi[N], di[N], ne[N], cfi[N], cdi[N], cne[N], va[N], sta[N], dis[N], dfn[N], low[N], bl[N], bar[N], ru[N], q[N], su[N], l, lc, ti, tp, st, SCC;
bool v[N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline void add(int x, int y)
{
	di[++l] = y;
	ne[l] = fi[x];
	fi[x] = l;
}
inline void add_c(int x, int y)
{
	cdi[++lc] = y;
	cne[lc] = cfi[x];
	cfi[x] = lc;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
inline int maxn(int x, int y)
{
	return x > y ? x : y;
}
void tarjan(int x)
{
	int i, y;
	dfn[x] = low[x] = ++ti;
	sta[++tp] = x;
	v[x] = 1;
	for (i = fi[x]; i; i = ne[i])
		if (!dfn[y = di[i]])
		{
			tarjan(y);
			low[x] = minn(low[x], low[y]);
		}
		else
			if (v[y])
				low[x] = minn(low[x], dfn[y]);
	if (!(dfn[x] ^ low[x]))
	{
		SCC++;
		do
		{
			y = sta[tp--];
			bl[y] = SCC;
			su[SCC] += va[y];
			v[y] = 0;
		} while (x ^ y);
	}
}
void spfa()
{
	int i, x, y, head = 0, tail = 1;
	q[1] = bl[st];
	dis[bl[st]] = su[bl[st]];
	while (head ^ tail)
	{
		x = q[++head];
		v[x] = 0;
		for (i = cfi[x]; i; i = cne[i])
		{
			y = cdi[i];
			if (dis[y] < dis[x] + su[y])
			{
				dis[y] = dis[x] + su[y];
				if (!v[y])
				{
					q[++tail] = y;
					v[y] = 1;
				}
			}
		}
	}
}
int main()
{
	int i, k, n, m, x, y, ma = 0;
	n = re();
	m = re();
	for (i = 1; i <= m; i++)
	{
		a[i].x = re();
		a[i].y = re();
		add(a[i].x, a[i].y);
	}
	for (i = 1; i <= n; i++)
		va[i] = re();
	st = re();
	k = re();
	for (i = 1; i <= k; i++)
		bar[i] = re();
	for (i = 1; i <= n; i++)
		if (!dfn[i])
			tarjan(i);
	for (i = 1; i <= m; i++)
	{
		x = bl[a[i].x];
		y = bl[a[i].y];
		if (x ^ y)
		{
			add_c(x, y);
			ru[y]++;
		}
	}
	spfa();
	for (i = 1; i <= k; i++)
		ma = maxn(ma, dis[bl[bar[i]]]);
	printf("%d", ma);
	return 0;
}